What's the deal on infinitesimal operators?

[bolbteppa]

It seems to me that if I try put this in a modern context, that I should avoid the idea of "a real number so close to another real number that it doesn't matter". Is there any way to look at it as an iterated limit? - along the lines of:

Limit as something ( perhaps \delta x, \delta y?)) approaches 0 of ( Limit as n -> infinity of a series of terms that are a function of alpha and the something).
Swap the limits.
= limit as n->infinity of ( limit of each term as the something ->0)
The limit of each term is an appropriate term for a power series in alpha
= Limit as n->infinity of a Taylor series in alpha.

Sorry, I should have explained the whole point of this when I said the Lie series for x_1 converges in my last post. If you assume x_1 = x_0 + Ux|_{x_0} \delta \alpha + ... converges then you have shown that x_1 can be expressed in terms of the infinitesimal generator \varepsilon(x_0,y_0) = \frac{\partial \phi}{\partial \alpha}(x_0,y_0,\alpha)|_{\alpha = \alpha_0} because the U^nx|_{x_0}'s when worked out only involve \varepsilon (x_0,y_0) (as in, just above equations 2.18 in Emanuel). What does this all mean? It means we can generate our one-parameter group of transformations T(x_0,y_0,\alpha) = (x_1,y_1) using infinitesimal transformations of the form \varepsilon(x_0,y_0) = \frac{\partial \phi}{\partial \alpha}(x_0,y_0,\alpha)|_{\alpha = \alpha_0} because we've shown that x_1 can be expressed in terms of the \varepsilon(x_0,y_0) = \frac{\partial \phi}{\partial \alpha}(x_0,y_0,\alpha)|_{\alpha = \alpha_0} to the first power. Thinking about it this does put transformations T in one-one correspondence with transformations of the group generated by the infinitesimal generator... The only issue I see as regards limits would be as to what your radius of convergence would be so as to also allow for approximations (like I'll mention below), though I might be wrong there, so that would be an interesting discussion but it's in the realm of Taylor series error approximation & nothing crazier, as far as I can see (correct me if I'm wrong, which is likely).

Therefore Emanuel's example on page 14 just below equations 2.18 of him starting with the infinitesimal generator - y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y} & ending up with the global one-parameter group of transformations (x_1,y_1) = (x_0\cos(\alpha) - y_0\sin(\alpha),x_0\sin(\alpha) + y_0 \cos(\alpha)) makes perfect sense. If you try the other method you've mentioned you'll end up with a completely wrong answer in this simple example (try it).

More can be said on this, we can use this idea to show how every transformation group is isomorphic to a translation group thus establishing the whole integration constant as our symmetry group when integrating differential equations, but there's no point in doing that until this makes sense.

Emmanuel remarks that he isn't concerned with the radius of convergence of the Taylor series. I don't know why. My intuition confused by the fact that the Taylor series expansion in alpha makes no (overt) assumption that alpha is small. Yet deriving seems to depend delicately on making (x1,y1) close to (x0,y0). I don't understand why(x0,y0) is needed at all. From some of what you just wrote, I have the thought that the Taylor series is first approached as a series in \delta x , \delta y and that the \alpha gets in there by a substitution.

Yeah your intuition is right, the issue about this being small is to ensure convergence of the global group equations we derive from the infinitesimal generator. I do not know the details of this, but if you look at the example of generating the infinitesimal generator of the rotation group given in Cohen on page 8 you see there comes a point where they derive the infinitesimal generator by assuming \alpha is close to zero so that \cos(\alpha) = 1, similarly for \sin(\alpha) = \delta \alpha. This is the kind of thing they mean when \alpha is small, & the justification is that in doing this you end up showing you can actually re-derive the original one-parameter group of transformations via a Lie series (as is done in Emanuel).

I think my derivation begun the other thread will work. I don't think it will be a fluke if I show it utilizes the result of post #49 of this thread.

You are basing your derivation off differentiating the "dx" & "dy" in \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy, you may as well differentiate the plus sign Please go ahead with it & hopefully you'll see what I mean when it doesn't work out, try applying your method to the example Emanuel gave, of deriving the rotation group from the infinitesimal transformations (below equations 2.18 on page 14) as a test before you go any further, this will illustrate what I mean when I'm saying you will differentiate the "dy" & "dy" terms, you'll not only not make sense you'll get the wrong answer when you try to generate the global rotation group starting from it's infinitesimal transformations (Emanuel's example).

 

[Stephen Tashi]

Not that the original poster of a thread can ever direct it! - but here is my simplistic view of the situation.

1.The straightforward Taylor expansion for the function in question is:
f(T_x(x,y,\alpha),T_y(x,y,\alpha) =
\ f(x,y) + (D_\alpha( f(T_y(x,y,\alpha),T_y(x,y,\alpha))_|{\alpha = 0})\ \alpha + (\frac{1}{2!}\ D^2_\alpha( f(T_y(x,y,\alpha),T_y(x,y,\alpha))_|{\alpha = 0})\ \alpha^2 + ...
The coefficients of the powers of \alpha do not depend on \alpha.

2. Without saying that I know what the operator U or its powers are, the claim to be proven is:
f(T_x(x,y,\alpha),T_y(x,y,\alpha) =
\ f(x,y) +(U f(T_x(x,y,\alpha),T_y(x,y,\alpha)))\ \alpha + (\frac{1}{2!}\ U^2 f(T_y(x,y,\alpha),T_y(x,y,\alpha)))\ \alpha^2 + ...

Edit: Taking bolbteppa's suggestion in a subsequent post that I think of the concrete example of the rotation group, I think the claim is actually:
f(T_x(x,y,\alpha),T_y(x,y,\alpha) =
\ f(x,y) +(U f(x,y)\ \alpha + (\frac{1}{2!}\ U^2 (f(x,y))\ \alpha^2 + ...

So the evaulations of the coefficients take place at (x,y).

3. There are two possibilities to consider about 2. Either the coefficients of the powers of \alpha are themselves functions of \alpha or they are not.

If they are not, then the coefficients of the respective powers of \alpha in 2. should equal the respective coefficients in the Taylor expansion in 1.) I see no fancy-dan footwork with manifolds, infinitesimals, or otherwise that can avoid this conclusion. Corresponding coeffiencts may not look the same when you compute them initially, but you should be able to prove they are the same.

If the coefficients in 2.) are functions of \alpha then 2. is a much more complicated expression than a simple power series in \alpha The coefficients of the corresponding powers of \alpha don't have to match because those in 2. would not be constant with respect to \alpha while those in 1. are.

4. As best I can tell, the "infinitesimal elements" u_x, u_y are not functions of \alpha since they are defined as evaluating derivatives at the value \alpha = 0.

5. The old books use a notation for U that amounts to U = u_x f_x + u_y f_y. What they mean by f_x, f_y is unclear to me. If f_x means "the partial derivative of f with respect to its first argument" then U applied to f(T_x(x,y,\alpha),T_y(x,y,\alpha) ) is going to produce a result that is a function of \alpha because the partial derivatives of f will be evaulated at the arguments (T_x(x,y,\alpha),T_y(x,y,\alpha)).

Edit: However, if my revised opinion of the claim 2. is correct the operator U is applied to f(x,y) not f(T_x(x,y,\alpha), T_y(x,y,\alpha) ).

(This is a separate issue than the question of whether U^2 implies differentiating the factors u_x, u_y upon the second application of U.)

 

[Stephen Tashi]

You are basing your derivation off differentiating the "dx" & "dy" in \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy, you may as well differentiate the plus sign Please go ahead with it & hopefully you'll see what I mean when it doesn't work out, try applying your method to the example Emanuel gave, of deriving the rotation group from the infinitesimal transformations (below equations 2.18 on page 14) as a test before you go any further, this will illustrate what I mean when I'm saying you will differentiate the "dy" & "dy" terms, you'll not only not make sense you'll get the wrong answer when you try to generate the global rotation group starting from it's infinitesimal transformations (Emanuel's example).

I'm not differentiating any infinitesimals. I'm differentiation functions. For the rotation group

T_x(x,y,\alpha) = - x \sin \alpha + y \cos(\alpha)
u_x(x,y) = D_\alpha \ T_x(x,y,\alpha)_{|\ \alpha = 0} = - y
T_y(x,y,\alpha) = x \cos(\alpha) + y \sin(\alpha)
u_y(x,y) = D_\alpha \ T_y(x,y,\alpha)_{|\ \alpha = 0} = x

So the results are consistent with those at the bottom of page 14 in Emmanuel's book:

U x = u_x \frac{\partial x}{\partial x} + u_y \frac{\partial x}{\partial y}
= -y (1) + x (0) = -y

U^2 x = U( Ux) = U (-y) = u_x \frac{\partial (-y)}{\partial x} + u_y \frac{\partial (-y)}{\partial y}
= (-y)(0) + x (-1) = -x

For matrix groups u_x and u_y are linear functions of x and y. I think if we want an example to show a problem with interpreting U^2 as involving terms like \frac{\partial ^2 u_x}{\partial x \partial y} we need an example where such terms are nonzero.

Edit:Thinking about what you meant, I believe one of my problems (here and in the other thread) is thinking that U is applied to f(T_x(x,y,\alpha),T_y(x,y,\alpha)). Actually the power series applies the operator U to f(x,y).

 

[bolbteppa]

4. As best I can tell, the "infinitesimal elements" u_x, u_y are not functions of \alpha since they are defined as evaluating derivatives at the value \alpha = 0. [/itex]

Yeah, they are functions of x & y, though again the operator U does not act on them because it is derived under the assumption that x & y are fixed at a point, but then because this fixed point was arbitrary we can consider them as variables. This is made clear in the Taylor series I posted in my first post yesterday.

5. The old books use a notation for U that amounts to U = u_x f_x + u_y f_y. What they mean by f_x, f_y is unclear to me. If f_x means "the partial derivative of f with respect to its first argument" then U applied to f(T_x(x,y,\alpha),T_y(x,y,\alpha) ) is going to produce a result that is a function of \alpha because the partial derivatives of f will be evaulated at the arguments (T_x(x,y,\alpha),T_y(x,y,\alpha)).
...

Edit: However, if my revised opinion of the claim 2. is correct the operator U is applied to f(x,y) not f(T_x(x,y,\alpha), T_y(x,y,\alpha) ).

The problem here amounts to lack of familiarity with conventions of notation. In this case you should be writing f(T_x(x_0,y_0,\alpha), T_y(x_0,y_0,\alpha) ) so that when \alpha is not zero (or \alpha_o we can write f(T_x(x_0,y_0,\alpha), T_y(x_0,y_0,\alpha) ) = f(x,y) thus applying U to f makes perfect sense, when applied to f we find that we should write U|_{(x_0,y_0)}f, & note that because of the |_{(x_0,y_0)} we are not dealing with something that is a function of \alpha in the end.

Edit:Thinking about what you meant, I believe one of my problems (here and in the other thread) is thinking that U is applied to f(T_x(x,y,\alpha),T_y(x,y,\alpha)). Actually the power series applies the operator U to f(x,y).

I've been trying to find a link to a proof of Taylor's theorem for multivariable functions (or even the second derivative test) analogous to the way it's done in Thomas calculus but I can't find one. What we are doing is treating f(T_x(x,y,\alpha),T_y(x,y,\alpha)) as a function of one variable, \alpha or \delta \alpha whichever you prefer, then expanding this single variable function g(\alpha) out in a Taylor series. However by using the chain rule we inadvertently end up with a Taylor expression for f in two variables. If you know Taylor's theorem in two variables this should be no problem (if you don't I'll post an explicit proof & an example or two no problem so that we're on the same page). The proof offers a perfect example of a moment analogous to the above where you might be confused into differentiating the u_x terms (they will be written as \frac{dx}{dt} in the proof) but you'll see it's not done there & it should be clear from your derivation why you shouldn't do it. Note that this is the entire idea, no U is required, we really should forget about it as it's just notation. We may introduce the U notation if we wish, afterwards, but the idea is that it lessens confusion whereas right now it's only creating more confusion thus forget about it. This can all be solved by going back to the derivation of Taylor's theorem, something you should try to do. There's an interesting comment in Cohen about the U that is highly relevant:

Since Uf can be written when the infinitesimal transformation is known, & conversely, \delta x = \varepsilon(x,y)\delta \alpha, ... is known when Uf is given, Uf is said to represent \delta x = \varepsilon(x,y)\delta \alpha, ... . For convenience of language we'll speak of "the infinitesimal transformation Uf" instead of "the transformation represented by Uf." But it must be borne in mind that Uf is not a transformation, it is only representative of one.

A really really short way to sum all this up is that we are using the chain rule, repeatedly in a way that turns things into Taylor's theorem. The U operator acts on f with respect to the x & y variables in the function (the first half of the chain rule), then the rest of the chain rule is already taken care of by the coefficients inside the U operator already, thus it's just a notational shortcut for the chain rule. Please try re-reading my first post from yesterday & point out any step in it you find iffy or unclear I think I know where we're going with all this stuff now, but I'll hold off until I've cemented everything.

 

[jostpuur]

I took a closer look at the Taylor series question.

<br /> \phi(x,y,\alpha) = (e^{\alpha U}f)(x,y)<br />

<br /> \psi(x,y,\alpha) = f\big(T_x(x,y,\alpha),T_y(x,y,\alpha)\big)<br />

In posts #44 #46 #49 #50 we succeeded in proving that

<br /> \frac{\partial}{\partial \alpha}\psi(x,y,\alpha) = U\psi(x,y,\alpha)\quad\quad\quad (*)<br />

holds. By the assumption that the PDE has a unique solution (which hasn't been verified, but probably holds anyway), this implied \phi=\psi.

If we assume that the Taylor series converge we get

<br /> e^{\beta D_{\alpha}}\big(f(T_x(x,y,\alpha),T_y(x,y,\alpha))\big)\Big|_{\alpha=0}<br /> = f(T_x(x,y,\beta),T_y(x,y,\beta)) = (e^{\beta U}f)(x,y)<br />

based on what was proven earlier.

Could it be that you want to prove this in some other way, not using the PDE interpretation? It doesn't look impossible.

The operators \frac{\partial}{\partial\alpha} and U commute, because the weights in U do not depend on alpha, and everything is smooth so that the partial derivatives commute too. So the induction step

<br /> \frac{\partial}{\partial\alpha}U^n\psi(x,y,\alpha) = U^{n+1}\psi(x,y,\alpha)<br />

is clear. This implies that e^{\beta\frac{\partial}{\partial\alpha}} and e^{\beta U} do the same thing when you operate \psi with them. Assuming that the Taylor series converge we get

<br /> \psi(x,y,\beta) = e^{\beta\frac{\partial}{\partial\alpha}}\psi(x,y,0) = e^{\beta U}\psi(x,y,0) = e^{\beta U}f(x,y)<br />

This way the PDE interpretation was not used, but still the technical result (*) which was proven in #44 #46 #49 #50 was needed.

Do these follow from some axioms about the transformation?

I think they do.

I don't follow the entire argument yet, but I at least understand the above question.

IMO you should take a closer look. The keys to the Taylor series results are there.

I can't interpret (f\circ T(\alpha))(x,y). The two arguments (x,y) sitting outside the parentheses confuse me.

It should not be seen as confusing. For example, if f,g:\mathbb{R}\to\mathbb{R} are mappings, then also f+g is a mapping. This means that we can denote (f+g):\mathbb{R}\to\mathbb{R} and the mapping is defined by

<br /> (f+g)(x) = f(x) + g(x)<br />

Also f\circ g is a mapping, which is defined by

<br /> (f\circ g)(x) = f(g(x))<br />

This how parentheses are usually used.

 

[jostpuur]

I'm going to add some details. In the previous post I said that \frac{\partial}{\partial\alpha} and U commute. This can get tricky with some notations though. But just to be clear, here's the proof. Assume that \xi(x,y,\alpha) is some three parameter function. Then

<br /> \frac{\partial}{\partial\alpha} U \xi(x,y,\alpha)<br /> = \frac{\partial}{\partial\alpha}\Big(u_x(x,y)\frac{\partial \xi(x,y,\alpha)}{\partial x}<br /> + u_y(x,y)\frac{\partial \xi(x,y,\alpha)}{\partial y}\Big)<br />
<br /> = u_x(x,y)\frac{\partial^2\xi(x,y,\alpha)}{\partial\alpha\partial x}<br /> +u_y(x,y)\frac{\partial^2\xi(x,y,\alpha)}{\partial\alpha\partial y}<br />
<br /> = u_x(x,y)\frac{\partial}{\partial x}\Big(\frac{\partial \xi(x,y,\alpha)}{\partial\alpha}\Big)<br /> + u_y(x,y)\frac{\partial}{\partial y}\Big(\frac{\partial \xi(x,y,\alpha)}{\partial\alpha}\Big)<br /> =U\frac{\partial}{\partial\alpha}\xi(x,y,\alpha)<br />

So they commute. At least here. Then we can substitute \xi=U^{n-1}\psi, \xi=U^{n-2}\psi and so on, when commuting \frac{\partial}{\partial\alpha} and U^n.

The stuff gets tricky if we are operating on the expression

<br /> f(T_x(x,y,\alpha),T_y(x,y,\alpha))<br />

If you take the derivative with respect to alpha (like operating with D_{\alpha} or \frac{d}{d\alpha}), the partial derivatives \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y} will appear with some additional factors, again depending on all three parameters. So do the operators commute now too?

I can admit I feel little confused by some of these questions. This is why I defined the functions \phi,\psi. They enabled me to avoid the confusion and ambiguities.

The fact is that partial derivatives are not well defined, if we are not clear about what the functions are, and what their parameters are. So operating with \frac{\partial}{\partial x} isn't neccessarily allowed, if we don't know the function on right. Consequently, operating with U isn't allowed always then either.

 

[Stephen Tashi]

One approach to proving the expansion

[eq 9.1]
f(T_x(x,y,\alpha),T_y(x,y,\alpha) = f(x,y) + U f(x,y)\ \alpha + \frac{1}{2!} U^2\ \alpha^2 + ....

using "just calculus" is to define U so that U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha) is (exactly) equal to D^n_\alpha f( T(_x(x,y,\alpha), T_y(x,y,\alpha)).

That will ensure
[eq. 9.2]
(U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0 } = (D^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0}.

The only question left open will be whether
[eq. 9.3]
(U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0 } = U^n f(x,y)

Settling that question may or may not require more advanced mathematics.

I'll summarize this approach (and ignore criticism that I use too many letters, in view of the fact that alternatives feel free to employ \phi,\psi,\eta,\xi, U, u_x, u_y,f,f_1, x, y, x_0,y_0,x_1,y_1,\epsilon, \delta x, \delta y, \delta \alpha.)The elements of the 1-parameter group are the mappings T(C): (A,B) \rightarrow ( T_x(A,B,C), T_y(A,B,C)).

We assume the parameterization with C is done so that

[Eq. 9.4]
T(0) = (A,B) (i.e. T(0) is the identiy map.)

[Eq. 9.5]
T( T(\beta), \alpha) = T(\beta + \alpha).

Assume all derivatives mentioned in this post exist.

The following two results were proved in post #49:

Theorem 9-1:

[Eq. 9.6]
\frac{\partial T_x}{\partial C} =\frac{\partial T_x}{\partial C}_{|\ C = 0}\frac{\partial T_x}{\partial A} + \frac{\partial T_y}{\partial C}_{|\ C = 0}\frac{\partial T_x}{\partial B}
[Eq. 9.7]
\frac{\partial T_y}{\partial C} =\frac{\partial T_x}{\partial C}_{|\ C = 0}\frac{\partial T_y}{\partial A} + \frac{\partial T_y}{\partial C}_{|\ C = 0}\frac{\partial T_y}{\partial B}

To prove theorem 9-1, let C = \alpha + \beta, differentiate both sides of the coordinate equations implied by T(\alpha + \beta) = T( T(\alpha),\beta) with respect to \beta. Set \beta = 0. Then deeply contemplate the verbal interpretation of the notation in the result!

Develop condensed notation for Theorem 9-1 by making the definitions:

[Eq. 9.8]
u_x(A,B) = \frac{\partial T_x}{\partial C}_{|\ C = 0}
[Eq. 9.9]
u_y(A,B) = \frac{\partial T_y}{\partial C}_{|\ C = 0}

[Eq. 9.10]
U_x(A,B,C) = u_x \frac{\partial T_x}{\partial A} + u_y \frac{\partial T_x}{\partial B}
[Eq 9.11]
U_y(A,B,C) = u_x \frac{\partial T_y}{\partial A} + u_y \frac{\partial T_y}{\partial B}

With that notation Theorem 9-1 amounts to:

[Eq. 9.12]
\frac{\partial T_x}{\partial C} = U_x
[Eq 9.13]
\frac{\partial T_y}{\partial C}= U_y

Define the differential operator U acting on a real valued function f of two variables (S,W) by:

[Eq. 9.14]
U f = U_x \frac{\partial}{\partial S} + U_y \frac{\partial}{\partial W}

An equivalent definition of U for a matrix group might be written in a simpler manner. In a matrix group, T_x and T_y are linear in A and B. Operations such as \frac{\partial T_x}{\partial A} or \frac{\partial T_y}{\partial B} will "pick-off" the functions of C that are the coefficients of A and B. (For example, T_x(x,y,\alpha) = x \cos(\alpha) - y \sin(\alpha),\ \frac{\partial T_x}{\partial x} = cos(\alpha).) For matrix groups, it may be simpler to write a definition of U that specifies the functions that are "picked off" directly by stating them as a matrix rather than doing this implicitly via the above definitions of U_x, U_y.

For U to be well defined, we must have specified a particular 1-parameter group since its definition depends on the definitions of U_x, U_y. The definition of U does not enforce any relationship between the variables S,W and the variables A,B,C involved with the group.

The most important application of U will be in the particular case when S = T_x(A,B,C), W=T_y(A,B,C).

Theorem 9-2: Let f(S,W) be a real valued function with S = T_x(A,B,C), W=T_y(A,B,C). Then

[Eq. 9.15]
\frac{\partial f}{\partial C} = U f

This proven by using the chain rule, theorem 9-1 and the definitions relating to U above.

If f(S,W) is understood to be evaluated at S = T_x(A,B,C), W=T_y(A,B,C) then functions derived from it such as \frac{\partial f}{\partial S} or \frac{\partial f}{\partial C} are understood to be evaluated at the same values. Hence Theorem 9-2 applies to them also and so we have results such as:

[Eq. 9.16]
\frac{\partial}{\partial C} \ (\frac{\partial f}{\partial S}) = U \ \frac{\partial f}{\partial S}

[Eq. 9.17]
\frac{\partial^2 f}{\partial C^2} = \frac{\partial}{\partial C} \ (\frac{\partial f}{\partial C}) = U \ \frac{\partial f}{\partial C} = U (U f)

Using the values A = x, B= y, C =\alpha the above reasoning shows (at least informally) that for any given integer n &gt; 0

[Eq 9.18]
D^n_\alpha f(T_x(x,y,\alpha),T_y(x,y,\alpha)) = U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha))

We now come to the question of how to prove eq 9.3, which said
(U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0} = U^n f(x,y)
(and also, whether it is actually true for all 1-parameter groups!)

I can't decide whether this result requires some sophisticated continuity argument about differential operators or whether result follows just from verbally interpreting the notation and definitions involved with U.

We can contrast, the meanings of:

[expression 9-A]
U^n\ f(x,y)

[expression 9-B]
(U^n\ ( f(T_x(x,y,\alpha),T_y),T_y(x,y,\alpha)))_{|\ \alpha = 0}

Using eq, 9.4, we can rewrite f(x,y) so expression 9-A becomes:

[expression 9-C]
U^n\ f( T_x(x,y,0), T_y(x,y,0))

This establishes that the variables in the definition of U^n f(x,y) are S = T_x(A,B,C), W = T_x(A,B,C), A = x, B = y, C = 0 As I interpret 9-C, it means that the relationships among the variables are enforced (including the fact that C = 0) and then the operator U^n is applied to f(S,W).

My interpretation of expression 9-B is that the relationships A = x, B = y , C =\alpha, S = T_x(A,B,C), W= T_y(A,B,C) are enforced. The operation of U^n is applied to f(S,W). Then C is set to zero after all that is done.

The question of whether expression 9-C equals expression 9-B hinges on whether the operation of setting C = 0 commutes with the operation of applying U^n.

I don't know much about differential operators, but I'll speculate that proving 9.3 by a continuity property of operators would involve something like showing:
(U^n(f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0} = \lim_{\alpha \rightarrow 0} U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha))
= U^n (lim_{\alpha \rightarrow 0} f(T_x(x,y,\alpha),T_y(x,y,\alpha)) = U^n f(T_x,(x,y,0),T_y(x,y,0)) = U^n f(x,y)

The significant feature of how U has been defined is that it does not involve any differentiations with respect to C. If there is a simple proof of 9.3 by parsing definitions, it would involve the claim that all functions involved in U^n, such as U_x(A,B,C), U_y(A,B,C), f(T_x(A,B,C),T_y(A,B,C)) and their various partial derivatives with respect to A, B give the same results, whether you first substitute C = 0 and perform the differentiations with respect to A, B or whether you do the differentiations first and then substitute C = 0.

 

[jostpuur]

Is there still a problem with this? I already thought that everything was reasonably solved.

 

[Stephen Tashi]

Is there still a problem with this? I already thought that everything was reasonably solved.

My goal is to define U explicitly as a differential operator. ( I don't know why definitions of U need to be so indirect and obfuscated. If U is differential operator then one idea for defining it is to define it in terms of differentiation - or is that too far out?)

I don't think definition of U as a differential operator that I offered in post #6 works in general, so I have proposed a different one. The definition I proposed in post #6 when applied to f( T_x(x,y,\alpha), T_y(x,y,\alpha) ) involves differentiating f with respect to its first and second arguments, but does not imply the factors that come from differentiating T_x and T_y with respect to their first and second arguments. Perhaps in your work, you were already using the definition that I propose in post #67.